Question

If x+
$\frac{1}{x}$
=5,Find the value of
${x}^{2}$
+
$\frac{1}{ {x}^{2} }$

Do it with full explanation
​

Answer 1

Answer:

Use the identity

$(x + \frac{1}{x})^{2} = {x}^{2} + { \frac{1}{x} }^{2} + 2 \\ {5}^{2} = {x {}^{2} + { \frac{1}{x} }^{} }^{2} + 2 \\ 25 - 2 = {x}^{2} + \frac{1}{x}^{2} \\ {x {}^{2} + { \frac{1}{x} }^{} }^{2} = 23$

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Answer 2

Step-by-step explanation:

Given :-

$\displaystyle \sf \dashrightarrow \: x + \frac{1}{x} = 5$

To Find :-

$\displaystyle \sf \dashrightarrow \: {x}^{2} + \frac{1}{ {x}^{2} }$

Solution :-

★ Squaring both sides,

$\displaystyle \sf \dashrightarrow \: {(x)}^{2} + \bigg( \frac{1}{x} \bigg) {}^{2} = {(5)}^{2}$

Using, (a + b)² = a² + 2ab + b²

$\displaystyle \sf \dashrightarrow \: {x}^{2} + 2 \times {x}^{2} \times \frac{1}{ {x}^{2} } + \frac{1}{ {x}^{2} } = 25$

$\displaystyle \sf \dashrightarrow \: {x}^{2} + 2 \times \cancel{{x}^{2} } \times \frac{1}{ \cancel{{x}^{2} }} + \frac{1}{ {x}^{2} } = 25$

$\displaystyle \sf \dashrightarrow \: {x}^{2} + 2 + \frac{1}{ {x}^{2} } = 25$

$\displaystyle \sf \dashrightarrow \: {x}^{2} + \frac{1}{ {x}^{2} } = 25 - 2$

$\displaystyle \sf \dashrightarrow \: {x}^{2} + \frac{1}{ {x}^{2} } = 23$

Hence,

$\displaystyle\dashrightarrow \: \underline{\boxed{ \bf{x}^{2} + \frac{1}{ {x}^{2} } = 23}}$